Problem: Solve for $x$ and $y$ using elimination. $\begin{align*}-2x-8y &= 1 \\ -x+6y &= 3\end{align*}$
We can eliminate $x$ when its corresponding coefficients are negative inverses. Recalling our knowledge of least common multiples, multiply the top equation by $-1$ and the bottom equation by $2$ $\begin{align*}2x+8y &= -1\\ -2x+12y &= 6\end{align*}$ Add the top and bottom equations. $20y = 5$ Divide both sides by $20$ and reduce as necessary. $y = \dfrac{1}{4}$ Substitute $\dfrac{1}{4}$ for $y$ in the top equation. $-2x-8( \dfrac{1}{4}) = 1$ $-2x-2 = 1$ $-2x = 3$ $x = -\dfrac{3}{2}$ The solution is $\enspace x = -\dfrac{3}{2}, \enspace y = \dfrac{1}{4}$.